GATE 2024 EC Q.31: solution and answer

GATE 2024 · ECQ.31 · 1 mark · NAT

In the given circuit, the current $I_{x}$ (in mA) is ______.

GATE 2024 EC question 31: A multi-branch circuit with a common bottom (ground) rail and a top rail split into two nodes, A (left) and B (right). — Figure from the GATE 2024 EC paper. It is part of the question.

Answer and solution

ANSWER

$I_{x} = 2 mA$

Official answer: 2Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

Take the bottom rail as ground. Call the left top node A and the right top node B. Ohm's law on the left 1 kΩ gives $I_{0} = \frac{V _{A}}{1 k Ω}$ , so with $V_{A}$ in volts, $I_{0} = V_{A}$ in mA.
The dependent source is a CCVS: its coefficient 1000 has units of ohms, so $1000 I_{0}$ is a voltage. Here $1000 I_{0} = (1000 Ω) \cdot \frac{V _{A}}{1000 Ω} = V_{A}$ . With − at A and + at B: $V_{B} = V_{A} + 1000 I_{0} = V_{A} + V_{A} = 2 V_{A}$ .
Ohm's law on the right 1 kΩ: $I_{x} = \frac{V _{B}}{1 k Ω} = 2 V_{A}$ (in mA, with $V_{A}$ in volts).
The middle branch is a 1 kΩ resistor in series with a 2 mA source, so it carries exactly $2 mA$ downward. A resistor in series with a current source does not change the branch current.
Apply KCL to the supernode that encloses A, B and the dependent source. Current in equals current out: $5 = I_{0} + 2 + I_{x} = V_{A} + 2 + 2 V_{A} (mA)$
Solve: $3 V_{A} = 3 \Rightarrow V_{A} = 1 V$ , so $I_{0} = 1 mA$ and $V_{B} = 2 V$ .
Hence $I_{x} = \frac{2 V}{1 k Ω} = 2 mA$ . Check: $1 + 2 + 2 = 5 mA$ , which equals the source current.

KEY IDEA

A floating dependent voltage source between two non-reference nodes calls for a supernode. Write KCL once around both nodes and use the source to link the node voltages ( $V_{B} = V_{A} + 1000 I_{0} = 2 V_{A}$ ). A current source in series with a resistor fixes that branch current at 2 mA, whatever the resistor.

WATCH OUT FOR

(1) Reversing the dependent source's polarity, i.e. writing $V_{B} = V_{A} - 1000 I_{0} = 0$ . That gives $I_{x} = 0$ and $V_{A} = 3 V$ . (2) Putting $I_{0}$ into $1000 I_{0}$ in mA instead of amperes, which inflates the source voltage by a factor of 1000. (3) Thinking the 1 kΩ in series with the 2 mA source changes that branch's current, or trying to write a KVL equation across a current source.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2024 EC Q.31: solution and answer

Answer and solution

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